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Precalculus Chapter 1 Section 4

Precalculus Chapter 1 Section 4. Building Functions from Functions. Combining Functions Algebraically. Let f and g be 2 functions with intersecting domains. Then for all values of x in the intersection, the algebraic combinations of f and g are defined by the following rules:

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Precalculus Chapter 1 Section 4

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  1. PrecalculusChapter 1 Section 4 Building Functions from Functions

  2. Combining Functions Algebraically • Let f and g be 2 functions with intersecting domains. Then for all values of x in the intersection, the algebraic combinations of f and g are defined by the following rules: • Sum: (f + g)(x)= f(x) + g(x) • Difference: (f – g)(x) = f(x) – g(x) • Product: (f g)(x) = f(x)g(x) • Quotient: (f/g)(x) = f(x)/g(x), provided g(x) ≠ 0 • In each case, the domain of the new function consists of all numbers that belong to both the domain of f and g.

  3. Combining Functions Algebraically • Example • Let f(x) = x2, and g(x) = √(x+1) • Find formulas for f + g, f – g, fg, f/g, & gg • First determine the domains of both f and g in order to determine the domain of the new function.

  4. Combining Functions Algebraically • Solution • f has a domain of all real numbers, g has a domain of x ≥ -1. • Find where these 2 domains intersect to find the domain of the new function.

  5. Combining Functions Algebraically • Composition of Functions • Let f and g be 2 functions such that the domain of f intersects the range of g. The composition f of g, denoted f ∘ g, is defined by the rule (f ∘ g)(x) = f(g(x)) • The domain of f ∘ gconsists of all x-values in the domain of g that map into g(x)-values in the domain of f.

  6. Combining Functions Algebraically • Example • Let f(x) = exand g(x) = √x . • Find (f ∘ g)(x) and (g ∘ f)(x) and verify numerically that the functions are not the same.

  7. Combining Functions Algebraically • Solution • (f ∘ g)(x) = f(g(x)) = f(√x) = e √x • (g ∘ f)(x) = g(f(x)) = g(ex) = √(ex) • Graph both on your calculator to verify they have different graphs.

  8. Combining Functions Algebraically • Finding the Domain of a Composition • Let f(x) = x2 - 1 & g(x) = √x . • Find the domains of • A) f ∘ g B) g ∘ f Solution: First find both compositions then determine the domain by looking at the domains of f and g individually.

  9. Combining Functions Algebraically • Finding the Domain of a Composition • A) • The domain of f(g(x)) is x ≥ 0, because √x must have non-negative values of x. • B) • The domain of g(f(x)) is x2 – 1 ≥ 0 , which means x2 ≥ 1, which means x ≥ 1 or x ≤ -1. • You can support this by looking at the graphs of each on your calculator.

  10. Modeling with Function Composition • In the medical procedure known as angioplasty, doctors insert a catheter into a heart vein and inflate a small, spherical balloon on the tip of the catheter. • Suppose the balloon is inflated at a constant rate of 44cubic millimeters per second. • Write an equation that relates volume (V) after t seconds. • Write an equation that relates volume (V) to its radiusr. • Write an equation that gives the radius r as a function of the time t. • What is the radius after 5 seconds?

  11. Modeling with Function Composition • Solution: After t seconds, V = 44t. • Solution: V(r) = 4/3πr3

  12. Modeling with Function Composition • Write an equation that gives the radius r as a function of the time. Solution: Solve the volume equation for r and substitute 44t for V. • What is the radius after 5 seconds? • After t = 5 seconds, the radius will be

  13. Relations and Implicitly Defined Functions • There are many useful curves in mathematics that fail the vertical line test and therefore aren’t functions. • For example, circles are very useful in the real world. • The equation for a circle, (x – h)2 + (y – k)2 = r2, is a relation but not a function of x. • A relation is the general term for a set of ordered pairs (x, y). • When we have equations that give us a relation but not a function, we can sometimes solve the equation for y and create 2 separate equations that yield 2 functions.

  14. Implicitly Defined Functions • Let’s look at the equation of a circle x2+y2=4 • While it is not a function itself, we can split it into two equations that do define functions, as follows We say that the relation given by the equation x2+y2=4 defines the 2 functions implicitly. They yield the upper and lower semicircles of the circle.

  15. Using Implicitly Defined Functions • Describe the graph of the relation x2 + 2xy + y2 = 1 • Solution: • Notice that the left side is a factorable trinomial. Use this to separate into 2 implicitly defined functions.

  16. Homework • Do problems # 1 – 8 all, # 11 – 29 odds, #31 – 43 odds, on pages 124 - 125

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